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(A) The power consumed by a series combination of bulbs is given by $P = \frac{V^2}{R_{eq}}$,where $V$ is the supply voltage and $R_{eq}$ is the equiv

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(A) The power consumed by a series combination of bulbs is given by $P = \frac{V^2}{R_{eq}}$,where $V$ is the supply voltage and $R_{eq}$ is the equivalent resistance of the series circuit.Since $V$ is constant,$P \propto \frac{1}{R_{eq}}$.When one bulb fuses,it is removed from the circuit. Since the bulbs are in series,the total resistance $R_{eq}$ of the circuit decreases $(R_{eq} = R_1 + R_2 + ... + R_n)$.As the total resistance $R_{eq}$ decreases,the total power consumed $P$ by the remaining bulbs increases.Therefore,the illumination in the room will increase.

Some electric bulbs are connected in series across a $220\, V$ supply in a room. If one bulb is fused,then the remaining bulbs are connected again in series across the same supply. The illumination in the room will

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